$\stackrel{\&conjugate0;}{x}\=\frac{1}{1\/6}{\int }_0^1\left({\left(\sqrt{y}\right)}^2-{\left(y\right)}^2\right)\/2 \mathit{\ⅆ}y$ = $1\/2$ 

$\stackrel{\&conjugate0;}{y}\=\frac{1}{1\/6}{\int }_0^{1}y \left(\sqrt{y}-y\right) \mathit{\ⅆ}y$ = $2\/5$ 

Calculate $A$, the area of region $R$ 

${\int }_0^1\left(\sqrt{y}-y\right) \mathit{\ⅆ}y$ = $\frac{1}{6}$$\stackrel{\text{assign to a name}}{\to }$$A$

Calculate $\stackrel{\&conjugate0;}{x}$ 

$\frac{1}{A}{\int }_0^1\left({\left(\sqrt{y}\right)}^2-{\left(y\right)}^2\right)\/2 \mathit{\ⅆ}y$ = $\frac{1}{2}$

Calculate $\stackrel{\&conjugate0;}{y}$ 

$\frac{1}{A}{\int }_0^{1}y \left(\sqrt{y}-y\right) \mathit{\ⅆ}y$ = $\frac{2}{5}$